An excess of potassium hydroxide solution was added to a solution containing 9.8 g of sulfuric acid.

An excess of potassium hydroxide solution was added to a solution containing 9.8 g of sulfuric acid. What mass of salt was formed in this case?

Let’s write the reaction equation:

2KOH + H2SO4 = K2SO4 + 2H2O

Let’s find the amount of sulfuric acid substance:

v (H2SO4) = m (H2SO4) / M (H2SO4) = 9.8 / 98 = 0.1 (mol).

According to the reaction equation, 1 mol of K2SO4 is formed per 1 mol of H2SO4, therefore:

v (K2SO4) = v (H2SO4) = 0.1 (mol).

Thus, the mass of the resulting potassium sulfate salt is:

m (K2SO4) = v (K2SO4) * M (K2SO4) = 0.1 * 174 = 17.4 (g).

Answer: 17.4 g.